# Math Help - Limit of a function that has a common denominator

1. ## Limit of a function that has a common denominator

lim as x goes to 1 of

(x^3 - 1)/(x^4 - 1)

Evaluated as shown above, 0/0 is obtained, which leads me to believe there is a common factor between the divisor and the dividend that can be cancelled so the limit can be evaluated. The problem is, I have no idea how to deal with difference of cubes and difference of 4th degrees. Any help would be much appreciated.

2. Originally Posted by evankiefl
lim as x goes to 1 of

(x^3 - 1)/(x^4 - 1)

Evaluated as shown above, 0/0 is obtained, which leads me to believe there is a common factor between the divisor and the dividend that can be cancelled so the limit can be evaluated. The problem is, I have no idea how to deal with difference of cubes and difference of 4th degrees. Any help would be much appreciated.
You can factor out $x-1$ and in both the numerator and the denominator, then cancel it:

$\frac{x^3-1}{x^4-1}=\frac{(x-1)\cdot(x^2+x+1)}{(x-1)\cdot (x^3+x^2+x+1)}=\ldots$

3. Originally Posted by evankiefl
lim as x goes to 1 of

(x^3 - 1)/(x^4 - 1)

Evaluated as shown above, 0/0 is obtained, which leads me to believe there is a common factor between the divisor and the dividend that can be cancelled so the limit can be evaluated. The problem is, I have no idea how to deal with difference of cubes and difference of 4th degrees. Any help would be much appreciated.
You could also use L'Hospital's Rule...

$\displaystyle \lim_{x \to 1}\frac{x^3 - 1}{x^4 - 1} = \lim_{x \to 1}\frac{\frac{d}{dx}(x^3 - 1)}{\frac{d}{dx}(x^4 - 1)}$

$\displaystyle = \lim_{x \to 1}\frac{3x^2}{ 4x^3}$

$\displaystyle = \frac{3}{4}$.

4. $\displaystyle\frac{{{x}^{3}}-1}{{{x}^{4}}-1}=\frac{(x-1)\left( {{x}^{2}}+x+1 \right)}{(x+1)(x-1)\left( {{x}^{2}}+1 \right)}.$

5. Thanks everyone, and interesting L'Hopital's Rule technique. What exactly is the rule?

6. With L'Hopital's Rule, whenever you have an indeterminant form (see below), you can take the derivative of the numerator, denominator, and then run the limit.

Indeterminant forms:

$\displaystyle\frac{0}{0}, \ \ \frac{\infty}{\infty}, \ \ 0*\infty, \ \ 1^{\infty}, \ \ 0^0, \ \ \infty^0, \ \ \infty - \infty$